Take into account that the period of a pendulum is given by the following formula:
[tex]T=2\pi\sqrt[]{\frac{l}{g}}[/tex]where,
l: length = 2.3 m
g: acceleration of gravity = ?
T: period
The period T is:
[tex]T=\frac{37s}{5.0}=7.4s[/tex]If you solve for g the equation of the period, you obtain:
[tex]\begin{gathered} T^2=4\pi^2\frac{l}{g} \\ g=\frac{4\pi^2l}{T^2} \\ g=\frac{4\pi^2(2.3m)}{(7.4s)^2}\approx1.66\frac{m}{s^2} \end{gathered}[/tex]Hence, the acceleration of gravity at the location of the pendulum is approximately 1.66m/s^2
